python解决线性代数问题
###转置
import numpy as np
import pandas as pd
a=np.array([[1,2,3],[4,5,6],[7,8,9]])
b=np.array([[1,2,1],[4,5,3],[8,1,6]])
print(a)
print(a.T)#A的转置
print(a.T.T)#A的转置的转置还是A本身[[1 2 3] [4 5 6] [7 8 9]] [[1 4 7] [2 5 8] [3 6 9]] [[1 2 3] [4 5 6] [7 8 9]]
##验证矩阵转置的性质：(A±B)'=A'±B'
print((a+b).T)
print((a.T+b.T))[[ 2 8 15] [ 4 10 9] [ 4 9 15]] [[ 2 8 15] [ 4 10 9] [ 4 9 15]]
#验证矩阵转置的性质：(KA)'=KA'
print((10*a).T)
print(10*(a.T))[[10 40 70] [20 50 80] [30 60 90]] [[10 40 70] [20 50 80] [30 60 90]]
print('a')
print(a)
print('b')
print(b)
print('a*b')
print((a*b))#位置相同相乘
#矩阵相乘
print(np.dot(a,b))#行列相称
a [[1 2 3] [4 5 6] [7 8 9]] b [[1 2 1] [4 5 3] [8 1 6]] a*b [[ 1 4 3] [16 25 18] [56 8 54]] [[ 33 15 25] [ 72 39 55] [111 63 85]]
方阵的迹就是主对角元素之和，使用trace()函数获得方阵的迹：
和的迹 等于 迹的和
print(a)
np.trace(a)[[1 2 3] [4 5 6] [7 8 9]]
Out[5]:
计算行列式
15
a=np.array([[1,2],[3,4]])
print(a)
print(np.linalg.det(a))[[1 2] [3 4]] -2.0000000000000004
求可逆矩阵 linalg.inv
np.linalg.inv(a)array([[-2. , 1. ],
[ 1.5, -0.5]])
np.linalg.solve()解一元线性方程组
# x + 2y = 3
# 4x ＋ 5y = 6
import numpy as np
A = np.mat('1,2; 4,5') # 构造系数矩阵 A
b = np.mat('3,6').T # 构造转置矩阵 b （这里必须为列向量）
r = np.linalg.solve(A,b) # 调用 solve 函数求解求特征值、特征向量。求矩阵的秩
x = np.diag((1, 6, 3))#创建一个对角矩阵！
print('x')
print(x)
a,b = np.linalg.eig(x)#特征值保存在a中，特征向量保存在b中
print('特征值a')
print(a)#特征值
print('特征向量b')
print(b)#特征向量
np.linalg.matrix_rank(x)x [[1 0 0] [0 6 0] [0 0 3]] 特征值a [1. 6. 3.] 特征向量b [[1. 0. 0.] [0. 1. 0.] [0. 0. 1.]]
求解方程
A = np.mat('1,5; 10,5') # 构造系数矩阵 A
print(A)
b = np.mat('4,8').T
print(b)
r=np.linalg.solve(A,b)
print(r)
# 构造转置矩阵 b （这里必须为列向量）
# 调用 solve 函数求解[[ 1 5] [10 5]] [[4] [8]] [[0.44444444] [0.71111111]]
用sympy包求解方程
from sympy import *
x = Symbol('x')
y = Symbol('y')
print(solve([10 * x - 3*y - 3, 3 * x + 4*y ],[x, y])){x: 12/49, y: -9/49}
x = Symbol('x')
solved_value=solve([x**2-100], [x])
print(solved_value)[(-10,), (10,)]
# 复数解
solved_value = solve([x ** 2 + 9], [x])
print(solved_value)
solved_value = solve([x ** 4 - 9], [x])
print(solved_value)[(-3*I,), (3*I,)] [(-sqrt(3),), (sqrt(3),), (-sqrt(3)*I,), (sqrt(3)*I,)]
#cos(x)-1=0
solved_value = solve([cos(x) - 0.5], [x])
print(solved_value)
solved_value = solve([cos(x) - 1], [x])
print(solved_value)[(1.04719755119660,), (5.23598775598299,)] [(0,), (2*pi,)]
# 二元二次方程组 x**3+6*x*y=0
# 3*x*y-2*x**2+3=0
x = Symbol('x')
y= Symbol('y')
solved_value=solve([x**2+6*x*y,3*x*y-2*x**2+3], [x,y])
print(solved_value)[(-sqrt(30)/5, sqrt(30)/30), (sqrt(30)/5, -sqrt(30)/30)]